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Question Number 32316 by NECx last updated on 23/Mar/18

$$Agasexpandsaccordingtothe$$$$lawpv=constant,wherepisthe$$$$pressureandvisthevolumeof$$$$thegas.Initially,v=1000m^{3}and$$$$p=40N/m^2.Ifthepressureis$$$$decreasedattherateof5{Nm}^{-2}{min}^{-1}$$$$findtherateatwhichthegasis$$$$expandingwhenitsvolumeis$$$$2000m^{3.}$$

Commented by NECx last updated on 23/Mar/18

$$Ididitthiswaybutthetextbook$$$$saysitswrong:$$$$$$$$pv=constant(p_1{v}_1=p_2{v}_2)$$$$\therefore 1000\times 40=2000\times p_2$$$$p_2=20{Nm}^{-2}$$$$\therefore dp=20-40=-20{Nm}^{-2}$$$$$$$$\Rightarrow \frac{dp}{dt}=5\Rightarrow dt=20/5=4min$$$$$$$$\therefore dv/dt=(2000-1000)/4$$$$=250m^3{min}^{-1}$$$$$$$$pleasecorrectmewhereIgotit$$$$wrong.$$

Commented by NECx last updated on 23/Mar/18

$$textbooksays500m^3{min}^{-1}$$

Commented by mrW2 last updated on 23/Mar/18

$$v=\frac{c}{p}$$$$\frac{dv}{dt}=-\frac{c}{p^2}\times \frac{dp}{dt}=-\frac{v}{p}\times \frac{dp}{dt}$$$$att=t_2:$$$$v=2000m^{3}andp=20N/m^2$$$$\Rightarrow \frac{dv}{dt}=-\frac{2000}{20}\times (-5)=500m^3/min$$$$$$$$\times \times \times \times \times \times$$$$p\left(t\right)=p_0-5t$$$$v\left(t\right)=\frac{c}{p_0-5t}$$$$v_0=\frac{c}{p_0}$$$$\Rightarrow v\left(t\right)=\frac{p_0{v}_0}{p_0-5t}=\frac{v_0}{1-\frac{5t}{p_0}}$$$$\frac{dv}{dt}\ne constant$$

Commented by NECx last updated on 23/Mar/18

$$thankyousomuch$$

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